/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 59 Moving a Nonsymmetric Triangular... [FREE SOLUTION] | 91影视

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Chapter 17: Problem 59

Moving a Nonsymmetric Triangular Pulse A long taut spring is started at a time \(t=0\) with a pulse moving in the \(+x\) direction in the shape given by the function \(f(x)\) with $$ f(x)=\left\\{\begin{array}{ll} \frac{1}{4} x+1 & -4

Short Answer

Expert verified
v = 10 m/s, u(x,t) = f(x-10t). Speed up portion: slope of f(x), similarly apply for down portion.

Step by step solution

01

Graph the Shape at t = 0

Draw the piecewise-defined function \[f(x)=\begin{cases} \frac{1}{4} x+1 & -4 1\), the function value is zero.
02

Calculate the Wave Speed

Use the formula \(v = \sqrt{\frac{F}{\mu}}\), where \(F = 5 \text{ N}\) and \(\mu = 50 \frac{g}{m} (0.05 \frac{kg}{m})\). Calculation yields \[v = \sqrt{\frac{5}{0.05}} = \sqrt{100} = 10 \frac{m}{s}. \]
03

Find the New Position at t = 0.001 s

Calculate the distance traveled over \( t = 0.001\) s with speed \(v = 10 \frac{m}{s}\): \[ d = v \cdot t = 10 \frac{m}{s} \cdot 0.001 s = 0.01 m = 1 cm. \] Shift the pulse rightwards by 1 cm and draw the graph.
04

Write the General Solution

The solution to the wave equation is a function of both position \(x\) and time \(t\). For this pulse moving rightwards, the solution is \[ u(x,t) = f(x - vt), \] where \(f(x)\) is the initial shape of the pulse and \(v = 10\ \text{m/s}\).
05

Calculate the Speed of a Segment of the String

During the upward motion before reaching the maximum displacement at any point on the string: Use \( v = \frac{dy}{dt}\) with the specific wave function. It is linear as per function segments.
06

Summary of Motion

While the pulse rises from 0 to peak, the speed is given by the slope of the pulse function. For descending, apply the same slope calculation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Wave Equation
The wave equation is a fundamental concept for understanding wave motion in strings. It is expressed mathematically as \[ \frac{\partial^2 u}{\partial t^2} = v^2 \frac{\partial^2 u}{\partial x^2} \]. This equation describes how waves propagate through a medium over time. For a string, the displacement of the string at any point \(u(x, t)\) depends on the position \(x\) and the time \(t\). This is crucial for predicting the movement of the wave. Understanding this equation helps to know how waves maintain their shape while traveling.
Wave Speed
In the context of a wave moving through a string, wave speed is calculated using the formula \[ v = \sqrt{\frac{F}{\mu}} \] where \(F\) is the tension in the string, and \(\mu\) is the mass per unit length of the string. For this problem, with \(F = 5\text{N}\) (tension) and \(\mu = 0.05 \frac{kg}{m}\) (mass density), we find \(v = 10 \frac{m}{s}\). This calculation is essential because the speed determines how quickly the wave moves along the string.
Piecewise Function
In our example, the displacement of the string is described by a piecewise function: \[f(x)=\begin{cases} \frac{1}{4}x+1 & -4
Displacement
Displacement in wave motion refers to how much a point on the string deviates from its equilibrium position as the wave passes. For our problem, the displacement at any time t is described by the function \(u(x,t) = f(x - vt)\), which shifts the initial pulse rightward over time. When drawing the graph of the wave at various points in time, we shift the entire wave form horizontally by an amount proportional to the time elapsed and the wave's speed. For instance, at \(t = 0.001\text{ s}\), the wave has moved 1 cm to the right.
Tension in String
Tension is the force exerted along the length of the string, which is a critical factor affecting wave propagation. In our case, the string is under a tension \(F\) of \(5 \text{ N}\). High tension typically results in higher wave speeds. The relationship, as derived in wave speed calculation, shows that greater tension leads to a faster wave (since \(v = \sqrt{\frac{F}{\mu}}\)). In practice, maintaining correct tension ensures the string vibrates predictably, allowing us to precisely describe movements through equations.

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Most popular questions from this chapter

Displacement of Particles A sinusoidal wave is traveling on a string with speed \(40 \mathrm{~cm} / \mathrm{s}\). The displacement of the particles of the string at \(x=10 \mathrm{~cm}\) is found to vary with time according to the equation \(y(x, t)=(5.0 \mathrm{~cm}) \sin [1.0 \mathrm{rad} / \mathrm{cm}-(4.0 \mathrm{rad} / \mathrm{s}) t] .\) The linear density of the string is \(4.0 \mathrm{~g} / \mathrm{cm} .\) What are the (a) frequency and (b) wavelength of the wave? (c) Write the general equation giving the transverse displacement of the particles of the string as a function of position and time. (d) Calculate the tension in the string.

Identical Except for Phase Two sinusoidal waves, identical except for phase, travel in the same direction along a string and interfere to produce a resultant wave given by \(y^{\prime}(x, t)=(3.0 \mathrm{~mm})\) \(\sin [(20 \mathrm{rad} / \mathrm{m}) x-(4.0 \mathrm{rad} / \mathrm{s}) t+0.820 \mathrm{rad}]\), with \(x\) in meters and \(t\) in seconds. What are (a) the wavelength \(\lambda\) of the two waves, (b) the phase difference between them, and (c) their amplitude \(Y\) ?

Show That Show that $$ \begin{array}{ll} y(x, t)=Y \sin k(x-v t), & y(x, t)=Y \sin 2 \pi\left(\frac{x}{\lambda}-f t\right) \\ y(x, t)=Y \sin \omega\left(\frac{x}{v}-t\right), & y(x, t)=Y \sin 2 \pi\left(\frac{x}{\lambda}-\frac{t}{T}\right) \end{array} $$ are all equivalent to \(y(x, t)=Y \sin (k x-\omega t)\).

Second Harmonic A rope, under a tension of \(200 \mathrm{~N}\) and fixed at both ends, oscillates in a second-harmonic standing wave pattern. The displacement of the rope is given by $$ y(x, t)=(0.10 \mathrm{~m})\left(\sin \left(\left(\frac{\pi}{2} \mathrm{rad} / \mathrm{m}\right) x\right) \sin ((12 \pi \mathrm{rad} / \mathrm{s}) t)\right. $$ where \(x=0.00 \mathrm{~m}\) at one end of the rope, \(x\) is in meters, and \(t\) is in seconds. What are (a) the length of the rope, (b) the speed of the waves on the rope, and (c) the mass of the rope? (d) If the rope oscillates in a third-harmonic standing wave pattern, what will be the period of oscillation?

Violin Strings The heaviest and lightest strings on a certain violin have linear densities of \(3.0\) and \(0.29 \mathrm{~g} / \mathrm{m} .\) (a) What is the ratio of the diameter of the heaviest string to that of the lightest string, assuming that the strings are of the same material? (b) What is the ratio of speeds if the strings have the same tension?
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